Tài liệu Physics exercises_solution: Chapter 23 docx

23.1:
J357.0
m150.0
1
m354.0
1
)C30.4)(C40.2(
11
12
21

















μμk
rr
qkqU
J.357.0 UW

23.2:


J104.5J109.1J109.1
888
ffi
UUUUW
J103.7
8

23.3: a)

.sm5.12
kg0015.0
J)491.0J608.0(2
2
1
J608.0
m800.0
)C1050.7)(C1080.2(
)sm0.22)(kg0015.0(
2
1
21
2
66
2








f
f
ffi
iii
v
r
qkq
mvEE
k
UKE
b) At the closest point, the velocity is zero:
m.323.0
J608.0
)C1080.7)(C1080.2(
J608.0
66
21




k
r
r
qkq
23.4:

.m373.0
J400.0
C)1020.7)(C1030.2(
J400.0
66
21





k
r
r
qkq
U
23.5: a)
.J199.0
m250.0
)C1020.1()C1060.4(
66




k
r
kQq
U

s.m6.37 ,J198.0)iii(
s.m7.36 J,189.0)ii(
s.m6.26
kg1080.2
J)0994.0(2
2
1
J0994.0
J0994.0
m5.0
1
m25.0
1
)C1020.1(C)1060.4(J0
(i) b)
4
2
66



















ff
ff
fff
fiif
vK
vK
vmvK
k
UUKK
23.6:
.J078.0C)102.1(66
m500.0
2
m500.0
262
22


kkq
kqkq
U
23.7: a)
J.1060.3
)m100.0(

)nC00.2)(nC00.3(

)m100.0(
)nC00.2)(nC00.4(
)m200.0(
nC)00.3)(nC00.4(
7
23
21
13
21
12
21





























k
r
qq
r
qq
r
qq
kU
.0,0b)
12
3231
12
21











xr
qq
x
qq
r
qq
kUIf
So solving for x we find:
.m360.0,m074.006.12660
2.0
68
600
2



xxx
xx
Therefore
m074.0x
since it is the only value between the two charges.
23.8: From Example 23.1, the initial energy
i
E
can be calculated:
J.1009.5

m10
)C1020.3)(C1060.1(

)sm1000.3)(kg1011.9(
2
1
19
10
1919
2631








i
iii
E
k
UKE
When velocity equals zero, all energy is electric potential energy, so:
.m1006.9
2
J1009.5
10
2
19


 r
r
ek
23.9: Since the work done is zero, the sum of the work to bring in the two equal charges
q must equal the work done in bringing in charge Q.
.
2
2
2
q
Q
d
kqQ
d
kq
WW
qQqq

23.10: The work is the potential energy of the combination.
J1031.7
3
2
2
m105
)C106.1()CNm100.9(
21
2
2
m105

m105
)2()(
m105
)(
m1025
)2(
19
10
219229
10
2
1010
10





































ke
eekekeeke
UUUU
epep
Since U is negative, we want do
J1031.7
19

to separate the particles
23.11:
12212211

so 0 ; UKUKUKUK 

eV9.00J1044.1
m1000.8with,
5
4
1221
4
18
1
10
2
00
2
1











U
r
r
e

πεrrrπε
e
U
23.12: Get closest distance
.γ
Energy conservation:
γ
ke
mvmv
2
22
2
1
2
1

m1038.1
)sm10()kg1067.1(
)C106.1()CNm109(
13
627
219229
2
2








mv
ke
γ
Maximum force:
N012.0
)m1038.1(
)C106.1()CNm109(
213
219229
2
2







γ
ke
F
23.13:
BBAA
UKUK 

sm42.72
J0.00550V)800V(200C)105.00(J00250.0)(
so,
6





mKv
VVqKK
qVKqVKqVU
BB
BAAB
BBAA
It is faster at B; a negative charge gains speed when it moves to higher potential.
23.14: Taking the origin at the center of the square, the symmetry means that the
potential is the same at the two corners not occupied by the
C00.5 μ
charges (The
work done in moving to either corner from infinity is the same). But this also means that
no net work is done is moving from one corner to the other.
23.15:
E

points from high potential to low potential, so
.and
ACAB
VVVV 
The force on a positive test charge is east, so no work is done on it by the electric
force when it moves due south (the force and displacement are perpendicular);
.
AD
VV 
23.16: a)

J.1050.1
6
 KqEdUW
b) The initial point was at a higher potential than the latter since any positive charge,
when free to move, will move from greater to lesser potential.
V.357nC)(4.20J)1050.1(
6


qUV
c)
C.N1095.5
m)06.0()nC20.4(
J1050.1
J1050.1
3
6
6





EqEd
23.17: a) Work done is zero since the motion is along an equipotential, perpendicular to
the electric field.
b)
J105.7)m670.0(
m
V

104.00nC)0.28(
44 








qEdW
c)
J1006.2)45cos60.2(
m
V
104.00nC)0.28(
34 








qEdW
23.18: Initial energy equals final energy:
s.m1089.6
J)102.88J1004.5(
kg1011.9

2
2
1
J1004.5
2
1
m0.40
C1000.2(
m0.10
C)10(3.00
C)1060.1(
J1088.2
m0.25
C)1000.2(
m0.25
C)10(3.00
C)1060.1(
2
1
6
1717
31
2
17
2
99
19
17
99
19

2
2
2
1
1
2
2
1
1









































f
fe
fef
i
fe
ffii
fi
v
vm
vmkE
kE

vm
r
keq
r
keq
r
keq
r
keq
EE
23.19: a)
m.105.2
V90.0
C)1050.2(
3
11





k
V
kq
r
r
kq
V
b)
m.105.7

V30.0
C)1050.2(
3
11





k
V
kq
r
r
kq
V
23.20: a)
C.1033.1
V)(48.0m)250.0(
9

kk
rV
q
r
kq
V
b)
V16
m)(0.750

C)1033.1(
9




k
V
23.21: a)
.V738
m0.05
C1050.6
m05.0
C1040.2
:AAt
99
2
2
1
1
























k
r
q
r
q
kV
A
b)
.V705
m0.06
C1050.6
m08.0
C1040.2
:BAt
99
2
2

1
1























k
r
q
r
q

kV
B
c)
J.108.25V)33(C)1050.2(
89 
 VqW
The negative sign indicates that the work is done on the charge. So the work done by the
field is
J.1025.8
8

23.22: a)
b)
.
4
1
2
0
a
q
πε
V 
c) Looking at the diagram in (a):
22
00
4
1
2
4
1

2)(
xa
q
πεr
q
πε
xV


d)
e) When
,
2
4
1
,
0
x
q
πε
Vax 
just like a point charge of charge
.2q
23.23: a)
b)
.0
)(




r
qk
r
kq
V
x
c) The potential along the x-axis is always zero, so a graph would be flat.
d) If the two charges are interchanged, then the results of (b) and (c) still hold.
The potential is zero
23.24: a)
.
2
)()(
:||
22
ay
kqy
ya
kq
ya
kq
Vay







.

2
)()(
:
.
2
)(
:
22
22
ay
kqa
ay
kq
ya
kq
Vay
ay
kqa
ay
kq
ya
kq
Vay















Note: This can also be written as













|||| ay
q
ay
q
kV
b)
c)
.
2

)()(
:
2
y
kqa
ay
kq
ya
kq
Vay






d) If the charges are interchanged, then the potential is of the opposite sign.
23.25: a)
b)
.
)(
)(
2
:
axx
axkq
ax
kq
x
kq

Vax






.
)(
)3(
2
:0
axx
axkq
xa
kq
x
kq
Vax





.
)(
)(
2
:0
axx

axkq
ax
kq
x
kq
Vx







Note: This can be also be written as
)(
||
2
||
ax
q
x
q
kV


c) The potential is zero at
3./and aax 
d)
e) For
,:

2
x
kq
x
kqx
Vax




which is the same as the potential of a point charge
–q. (Note: The two charges must be added with the correct sign.)
23.26:a)
.
2
||
12
||
22










ya

y
kq
r
kq
y
kq
V
b)
.
3
3
4
when,0
22
22
2
a
yay
ya
yV



c)
d)
,
21
:
y
kq

yy
kqVay










which is the potential of a point charge
q
.
23.27:
J.104.72C)1060.1()V295(
1719 
 VqUW
But also:
s.m1001.1
kg109.11
J)1072.4(2
2
1
7
31
17
2







vmvKW
23.28: a)
m.415.0
CN0.12
V98.4

E
V
d
d
V
E
b)
C.1030.2
)m415.0()V98.4(
10

kk
Vd
q
d
kq
V
c) The electric field is directed away from q since it is a positive charge.
23.29: a) Point b has a higher potential since it is “upstream” from where the positive

charge moves.
0)(||)(||)(  abEVVabEabEVV
abba
b)
C.N800
m3.0
240

V
d
V
E
c)
J.104.8V)240(C)1020.0(
56 
 VqUW
23.30:(a)
,0
2

QQ
VVV
so V is zero nowhere except for infinitely far from the
charges.
The fields can cancel only between the charges
22
22
2
2)(
)(

)2(
xxd
xd
Qk
x
kQ
EE
QQ



.
21

d
x
The other root,
,
21

d
x
does not lie between the charges.
(b)
V can be zero in 2 places, A and B.
.oflefttheto
0
)2()(
:
30

)2()(
:
2
QEE
dy
yd
Qk
y
Qk
Bat
dx
xd
Qk
x
Qk
Aat
QQ









12
)(
)2(
22





d
x
xd
Qk
x
kQ
(c)
Note that E and V are not zero at the same places.
23.31: a)
2211
qVKqVK 

;)(
1221
KKVVq 
C10602.1
19
q

J;10099.4
182
1e
2
1
1


 vmK J10915.2
172
2e
2
1
2

 vmK

V156
12
21



q
KK
VV
The electron gains kinetic energy when it moves to higher potential.
b) Now
0J,10915.2
2
17
1


KK

V182
12

21



q
KK
VV
The electron loses kinetic energy when it moves to lower potential
23.32: a)
.V6.65
m0.48
C)1050.3(
9




k
r
kq
V
b)
V3.131
m0.240
C)1050.3(
9





k
V
c) Since the sphere is metal, its interior is an equipotential, and so the potential
inside is 131.3 V.
23.33: a) The electron will exhibit simple harmonic motion for
,ax 
but will
otherwise oscillate between
.cm0.30
b) From Example 23.11,
V796
m)(0.150m)(0.300
1
m0.150
1
C)10(24.0
Δ
11
22
9
2222

























kV
ax
a
kQV
ax
kQ
V
But
s.m1067.1
2
1
7
kg109.11
V)(796C)1060.1(2
2

31
19





vmvVqW
23.34: Energy is conserved:
.V0117.0
C)1060.1(2
)sm1500()kg1067.1(
2
1
19
227
2






VVqmv
But:
m.158.0
m/C1000.5
)V0117.0(2
exp)m180.0(
λ

2
exp
λ
2
exp)ln(
2
λ
12
0
0
0
0
00
0





























πε
r
V
πε
rr
V
πε
rrrr
πε
V
23.35: a)
C.N8000
m0450.0
360

V
d
V

E
b)
N.101.92C)10(2.40)CN8000(
59 
 EqF
c)
J.108.64m)(0.0450N)1092.1(
75 
 FdW
d)
J.108.64C)1040.2()V360(
79 
 VqU
23.36: a)
V.18.2m)10(3.8)CN480(
2


EdV
b) The higher potential is at the positive sheet.
c)
.mC104.25)CN480(
29
0
0

 εσ
ε
σ
E

23.37:a)
m.1058.1
mV1000.3
V4750
3
6




E
V
d
d
V
E
b)
.mC1066.2)mV1000.3(
256
0
0

 εσ
ε
σ
E
23.38: a)
C.N5311
mC100.47
0

29
0




εε
σ
E
b)
V.117m)(0.0220C)/N5311(  EdV
c) The electric field stays the same if the separation of the plates doubles, while the
potential between the plates doubles.
23.39: a) The electric field outside the shell is the same as for a point charge at the center
of the shell, so the potential outside the shell is the same as for a point charge:
R.for
4
0
 r
r
πε
q
V
The electric field is zero inside the shell, so no work is done on a test charge as
it moves inside the shell and all points inside the shell are at the same potential as the
surface of the shell:
.for
4
0
Rr

R
πε
q
V


nC20
)V1200()m15.0(
sob)



kk
RV
q
R
kq
V
c) No, the amount of charge on the sphere is very small.
23.40: For points outside this spherical charge distribution the field is the same as if all
the charge were concentrated at the center.
Therefore
2
0
4 rπε
q
E

and
C1069.1

/N.m109
)m200.0()CN3800(
4
8
229
2
2
0




C
Erπεq
Since the field is directed inward, the charge must be negative. The potential of a point
charge, taking

as zero, is
V760
m200.0
)C1069.1()C/N.m109(
4
8229
0




rπε
q

V
at the surface of the sphere. Since the charge all resides on the surface of a conductor, the
field inside the sphere due to this symmetrical distribution is zero. No work is therefore
done in moving a test charge from just inside the surface to the center, and the potential at
the center must also be
.V760
23.41: a)
.VE 

.CyBxAxy
zz
V
E
C.AxCyBxAxy
yy
V
E
Bx.AyCyBxAxy
xx
V
E
z
y
x
0)(
)(
2)(
2
2
2






















b)










A
C
A
B
y
A
C
xCAxx
A
B
yBxAy
.
2
so0,
2
02









z
A
BC
A

C
E
A
BC
,
2
,at0,
2
22
23.42: a)
VE 
.
)(
323222
222
r
kQx
zyx
kQx
zyx
kQ
xx
V
E
x



















Similarly,
.and
33
r
kQz
E
r
kQy
E
zy

b) So from (a),
,
ˆ
ˆˆˆ
22
r

kji
r
kQ
r
z
r
y
r
x
r
kQ
E










which agrees with Equation (21.7).
23.43: a) There is no dependence of the potential on
,or yx
and so it has no
components in those directions. However, there is
z
dependence:
.0,

ˆ
dzforCEC
z
V
EVE
z



 k
and
,for,0 dz E

since the potential is constant there.
(b) Infinite parallel plates of opposite charge could create this electric field, where the
surface charge is
.
0
Cεσ 